Promoting, advancing and defending Intelligent Design via data, logic and Intelligent Reasoning and exposing the alleged theory of evolution as the nonsense it is.
I also educate evotards about ID and the alleged theory of evolution one tard at a time and sometimes in groups

Thursday, June 13, 2013

keiths, Total Moron

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Give up keiths, you are nothing but a lowlife moron. Now keiths sez:

we have asserted that the cardinality of a set does not depend on the identity of its elements — only on their numerosity.

Umm the numerosity depends on their identity on the number line.

To Joe, that means that we are claiming that “numbers have no meaning.”

They don't if you place them in a one-to-one correspondence by rearranging their position on that number line.

Substitute one number for another and the cardinality of a set doesn’t change: {1,2,3} has the same cardinality as {5,6,7}.

Hey lowlife, we are talking about allegedly infinite sets. Therefor bringing a finite set into the discussion is dishonest.

Then to prove he is an asshole, keiths sez:

So Mr. “I fix things- all kinds of things- mechanical, electrical, electronic and personal” was unable to connect to a website and download some PDFs without outside help?

LoL! I didn't need any outside help- I did it on my own. Just because people offered to help doesn't mean I needed it.

And keiths continues his cowardly lies:

It has never dawned on Joe that the process of counting — which he wholeheartedly endorses, even for infinite sets — is exactly the sort of thing he says should never be done.

Your false accusation means nothing. But it does expose you are a liar.

Why? Because it involves setting up a one-to-one correspondence between non-identical elements.

Counting does NOT require such a thing you lowlife moron. And not only that Jerad now agrees with me!

How do you count the elements in {5, 23, 41, 99, 666}?

By counting them, asshole. Again with the finite set as if your deception and dishoesty mean something.

And not to be outdone on the TARD, Richie the coward, chimes in:

If Joe had two beans and one had “15″ pained on it, he’d have 16 beans by Joemath.

If Richie and keiths had a brain, they would be dangerous. You assholes are pathetic and apparently you are proud of your cowardice.

I dare you two to try the following:

Try it- start at 0 and count every non-negative integer with one counter
and every positive even integer with another. The counter counting the
non-negative integers will always be at least 2x that as the other counter, ie
it will always have more elements- ALWAYS- as long as infinity exists and
especially when infinity ceases to exist.

Invisible ink- I should have tried that, hand in a blank paper and say the lines are there, they just don't have any width. And all the points are there, they just don't have any area.

While you can't draw a true geometric line or point, they are useful, as you demonstrated when you used the Pythagorean Theorem. There's also the 3-4-5 rule for constructing right angles. All that follows from working with "invisible" lines.

The problem is you can't just use the "original" proof. All the proofs I've ever seen assume (Euclidean) plane geometry. The theorems of Euclidean plane geometry do not all hold in JoeMath. For example, in Euclidean geometry, every line segment has a unique midpoint. Not so in JoeMath.

Also, any two distinct nonparallel lines intersect in exactly one point in Euclidean geometry. That is not the case in JoeMath.

Why wouldn't there be a single midpoint for every line segnebt in JoeMath? Please be specific.

Choose a line L and two adjacent points A and B on L (they're like beads on a string, right?). Let's say their coordinates are A = (x1, y1) and B = (x2, y2). Then the midpoint should be ( (x1 + x2)/2, (y1 + y2)/2 ), but those are not the coordinates of a point in JoeMath. The centers of any two points are at least one electron's diameter apart in your system.

LoL! I can't seem to keep my hands lined up and the timing correct without looking at the keyboard!

What the fuck!!!!!!!

Choose a line L and two adjacent points A and B on L (they're like beads on a string, right?). Let's say their coordinates are A = (x1, y1) and B = (x2, y2). Then the midpoint should be ( (x1 + x2)/2, (y1 + y2)/2 ), but those are not the coordinates of a point in JoeMath.

JoeMath says the midpoint between two adjacent points would be their touching boundary, ie exactly 1D into the line segment(from either end) consisting of two adjacent points. (where D = diameter of a point)

"JoeMath says the midpoint between two adjacent points would be their touching boundary, ie exactly 1D into the line segment(from either end) consisting of two adjacent points. (where D = diameter of a point)"

Try it- start at 0 and count every non-negative integer with one counter and every positive even integer with another. The counter counting the non-negative integers will always be at least 2x that as the other counter, ie it will always have more elements- ALWAYS- as long as infinity exists and especially when infinity ceases to exist.

And no one can demonstrate otherwise. All Jerad can do is act like the little whiny baby that he is.

And what I am proposing has nothing to do with the rate. One set will always have more elements tahn the otehr- always, forever, for infinity, even.

And if points don't have a dimension, ie no diameter, then how do you know they exist?

JoeMath says the midpoint between two adjacent points would be their touching boundary, ie exactly 1D into the line segment(from either end) consisting of two adjacent points. (where D = diameter of a point)

You could make one by using the midpoints of both points (A,B) as the diameter. Then the boundary of AB will be that point's midpoint.

That would be the "logical" thing to do, but you can't do it in JoeMath.

In your system, each line is a series of points lined up like beads on a string. This new midpoint wasn't one of those, so it can't be on the line.

On the other hand, in Euclidean geometry, the midpoint of any segment does lie on the line determined by the endpoints of the segment.

That means there are theorems in Euclidean geometry which are false in JoeMath. Therefore you can't just say "use the original proof of the Pythagorean Theorem" in your system, without checking whether the proof is valid.

BTW, there are hundred(s) of proofs of the Pythagorean Theorem, maybe you could look through them and try to find one that works in JoeMath? I don't think any of them will, but the proofs are interesting to read.

"And you do NOT get to say that a midpoint has to be an actual point."

Way down the rabbit hole now.

"In Euclidean geometry points do NOT exist. As I said any point without dimension is pointless."

It must be pretty dark and scary down there.

"IOW there isn't any "midpoint" because points do not exist!But anyway, go fuck yourself for trying to tell me what JoeMath can and cannot do."

Even Joe doesn't know what JoeMath can and cannot do. For instance he says that (0, 1) has a smallest value but he can't find it.

He says the cardinality of {1, ½, ⅓, ¼ . . . } is greater than the cardinality of {1, ½, ¼, ⅛ . . . } but he can't say what either of them are or how they compare with the cardinality of {1, 2, 3, 4 . . . . }. Somehow JoeMath thinks cardinality has to do with the kind of elements that are in the sets.

And JoeMath can't tell us what the legth of the diagonal of a unit square is.

Sure I can. Once you describe how your system works, anyone can start conducting research on JoeMath.

And you do NOT get to say that a midpoint has to be an actual point.

So if a midpoint is not a point, what is it? What are its dimensions? It almost looks as if you are thinking of midpoints as being locations without size, which is how Euclidean points are regarded.

In Euclidean geometry points do NOT exist. As I said any point without dimension is pointless.

False. If you are working with the usual x-y plane model for Eucldidean geometry, points are just ordered pairs of real numbers. Like (2, 3). If I asked you to find the distance between (2, 3) and (-1, 7), no doubt you would say 5. The fact that ordered pairs of real numbers are not little disks did not prevent us from doing the calculation.

BTW, Joe, all this leads to another question: Choose two lines L1 and L2 in JoeMath. Which line, if any, has more points? Answer: All lines are identical in JoeMath, so the cardinalities should be equal.

You can certainly choose a different word to use for "place" if you want to avoid confusion.

Anyway, this is how math works. It turns out that lines in JoeMath present certain problems. Midpoints, intersections of lines, and so forth. And what happens if next year they announce that the official diameter of the electron has been changed? Then your life's work is shot to hell.

It's much easier to dispense with these disk-like points and just say a line is a set of places. Using sequences, you could then extend the original set of places so that it is topologically equivalent to the real number line. This carries over to the x-y plane as well. And once you have the x-y plane, then you can prove the Pythagorean Theorem.

Seeing that you are too stupid to understand that the faster the rate of count means that more elements will be counted and that more elements means a greater cardinality, perhaps mathematics isn’t your thing and you just should shut up.

"Seeing that you are too stupid to understand that the faster the rate of count means that more elements will be counted and that more elements means a greater cardinality, perhaps mathematics isn’t your thing and you just should shut up."

Oh dear, is there an echo in here? Didn't I just read this on another thread? I'm pretty sure I did. Which means that not only are your methods not able to deliver but your ability to respond with intelligent and insightful reponses has also reached the end of the tracks. But, the inifinity train continues to chug along, to infinity and beyond!! Maybe, if you run really fast and count those positive integers all along the way, you can catch up. I'm not sure we can slow down the progression of mathematical thinking much more . . . so many people doing so much work . . . I guess if you can't keep up . . .

Seeing that you are too stupid to understand that the faster the rate of count means that more elements will be counted and that more elements means a greater cardinality, perhaps mathematics isn’t your thing and you just should shut up.

Seeing that you are too stupid to understand that the faster the rate of count means that more elements will be counted and that more elements means a greater cardinality, perhaps mathematics isn’t your thing and you just should shut up.

"Seeing that you are too stupid to understand that the faster the rate of count means that more elements will be counted and that more elements means a greater cardinality, perhaps mathematics isn’t your thing and you just should shut up."

I'm waiting to see you compare the cardinalities of the positive integers and {1, ½, ⅓, ¼ . . . .} since you're such a great counter. Go on. Or are you going to choke on this problem AGAIN?

" "Given 2 sets, A and B, if A contains all of the members of B AND has members B does not, A's cardinality has to be greater than B's."

And you sure as hell cannot demonstrate that is wrong."

Sure I can. I have done. I've shown there is a one-to-one correspondence between the positive integers and the positive even integers and you don't get the proof so you think you can say I haven't proved that.

You are not the final arbitrator. But you don't get that either. You think you get some kind of pass because . . . I don't know why you think that way. But you don't get a pass. And you're not right. And 100 years of mathematical work which you don't understand say so. And hundreds of mathematicians say so. All all of those people also know that there is no smallest element of (0, 1). Did you think I was going to forget? Sorry. And all of those people think that root 2 has an exact value that cannot be expressed with a finite decimal expansion. But you don't get it so you think you're right.

Prove me wrong, give me an exact value for a number which squared equals 2. There must be such a number. It's the solution for the equation x^2 = 2. Or is it, in JoeMaths, that that equation has no solution? Please answer.